1983 AIME Problems/Problem 9

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Problem

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$.

Solution

Let $y=x\sin{x}$. We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$.

Since $x>0$ and $\sin{x}>0$ because $0< x<\pi$, we have $y>0$. So we can apply AM-GM:

\[9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12\]

The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23$.

Therefore, the minimum value is $\boxed{012}$ (when $x\sin{x}=\frac23$; since $x\sin x$ is continuous and increasing on the interval $0 \le x \le \frac{\pi}{2}$ and its range on that interval is from $0 \le x\sin x \le \frac{\pi}{2}$, by the Intermediate Value Theorem this value is attainable).

See also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions